Arectangle is 8cm longer then it is wide, and it’s area is 153cm^2. find the dimensions of the rectangle.

Width of 9cm and length of 17cm.

Step-by-step explanation:

To calculate an area of a rectangle, use the formula A=l*w. We know the length is 8cm longer than the width. So l = 8 + w. So the area is A= (8+w)*w.

We also know the area is 153. Substitute this value for A and solve.

To solve the quadratic, use the quadratic formula:

Here a=1, b=8 and c=-153.

Since w=9 or w=-17, substitute this value for w in l= 8+w to find l.

l = 8+9 = 17

or

l=8+-17 = -9

Since length cannot be positive, it must be l=17 and w=9.

the line of

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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